If $f$ is Lebesgue integrable, then prove that $F$ is continuous and differentiable over $\mathbb{R}$. Assume that a function $f$ is integrable over $\mathbb{R}$ and define $$F(w) = \int_{\mathbb{R}} f(t)\sin(wt) dt,$$
for $w \in \mathbb{R}$. Prove that $F$ is continuous over $\mathbb{R}$, and that if the function $tf(t)$ is integrable over $\mathbb{R}$ then $F$ is differentiable over $\mathbb{R}$.
I am pretty sure we can solve this problem using Dominated Convergence Theorem. I have tried constructing a sequence of functions that will satisfy the hypothesis of DCT with the integrand of the function, here is what I have so far. Fixing $w$ define $h_n(t) = f(t)\sin((w+1/n)t) \rightarrow f(t)\sin(wt)$ for all $t$. Moreover $|h_n(t)| \leq |f(t)|$ and since $f$ is integrable then so is $|f|$. This satisfies both conditions of DCT so we have that $|\int (f(t)\sin(wt) - h_n)| \rightarrow 0$ as $n \rightarrow \infty.$ In other words we have shown that $$|F(w)-F(w+1/n)| \rightarrow 0,$$ as $n \rightarrow \infty$. Not too sure if this is valid, but please point out any mistakes that you may find. As for differentiability, I am not too sure how to go about it. We are given that $\int_{\mathbb{R}}|tf(t)| < \infty$, but I am not sure how to use it.
Thanks in advance!
 A: To show that $F$ is continuous, it suffices that $F$ is continuous
at an arbitrary point $w\in\mathbb{R}$.
Let $w\in\mathbb{R}$ be arbitrary. Let $(w_{n})$ be an arbitrary
sequence in $\mathbb{R}$ such that $w_{n}\rightarrow w$. We go to
prove that $F(w_{n})\rightarrow F(w)$. For each $n$, let $g_{n}:\mathbb{R}\rightarrow\mathbb{R}$
be defined by $g_{n}(t)=f(t)\sin(w_{n}t)$. Let $g:\mathbb{R}\rightarrow\mathbb{R}$
be defined by $g(t)=f(t)\sin(t)$. Clearly $|g_{n}|\leq|f|$ and $g_{n}\rightarrow g$
pointwisely. By Lebesgue Dominated Convergence Theorem, $\int g_{n}\rightarrow\int g$.
That is, $F(w_{n})\rightarrow F(w)$.

We go to prove that if $t\mapsto tf(t)$ is integrable, then $F$
is differentiable.
Let $w\in\mathbb{R}$ be fixed. We go to show that the derivative
$F'(w)$ exists. Let $(w_{n})$ be an arbitrary sequence such that
$w_{n}\neq w$ and $w_{n}\rightarrow w$. We go to show that the limit
$\lim_{n\rightarrow\infty}\frac{F(w_{n})-F(w)}{w_{n}-w}$ exists. By direct calculation,
for $t\neq0$,
\begin{eqnarray*}
 &  & f(t)\frac{\sin(w_{n}t)-\sin(wt)}{w_{n}-w}\\
 & = & [tf(t)]\frac{\cos\frac{(w_{n}+w)t}{2}\sin\frac{(w_{n}-w)t}{2}}{\frac{(w_{n}-w)t}{2}}.
\end{eqnarray*}
For each $n$, define $h_{n}:\mathbb{R}\rightarrow\mathbb{R}$ by
$h_{n}(t)=f(t)\frac{\sin(w_{n}t)-\sin(wt)}{w_{n}-w}.$ Recall that
for any $x\in\mathbb{R}\setminus\{0\}$, $\left|\frac{\sin x}{x}\right|\leq1$.
Therefore, $|h_{n}(t)|\leq|tf(t)|$. Also notice that for each $t\neq0$,
$\frac{\sin\frac{(w_{n}-w)t}{2}}{\frac{(w_{n}-w)t}{2}}\rightarrow1$
as $n\rightarrow\infty$. Hence, $h_{n}(t)\rightarrow[tf(t)]\cos(wt)$.
Since $t\mapsto tf(t)$ is integrable, by Lebesgue Dominated Convergence
Theorem, $\int h_{n}(t)dt\rightarrow\int tf(t)\cos(wt)dt$. On the
other hand, observe that $\int h_{n}(t)dt=\frac{F(w_{n})-F(w)}{w_{n}-w}$.
Hence, $\lim_{n\rightarrow\infty}\frac{F(w_{n})-F(w)}{w_{n}-w}=\int tf(t)\cos(wt)dt.$
This shows that $F'(w)$ exists and $F'(w)=\int tf(t)\cos(wt)dt$.
Remark: If we formally differentiate $F'(w)=\int f(t)\frac{d}{dw}\sin(wt)dt=\int tf(t)\cos(wt)dt$,
we get the same results.
A: Danny’s answer covers continuity. For differentiability, you may want to consider applying the Mean Value Theorem.
